# Combinatorics problem with “at least” condition

I had a regular combinatorcics exercise to solve and I thought it's possible to solve it in two ways but it turned out that only one way is correct. It is: A team of 4 students is to be selected for a competition. There are 8 boys and 12 girls to choose from. If the team have to include at least one boy and at least one girl, in how many ways can the team be selected?

So the simplest solution which is correct is to find the total number of possible selections (without any conditions) and then subtract from it the number of possible solutions including only boys and only girls.

But I thought the method mentioned below would be correct, it turned out to be wrong: (8 nCr 1)(12 nCr 1)(18 nCr 2) = 14688 The two first pair of parantheses mean that we need at least one boy and at least one girl of our group of 20. And (18 nCr 2) means that we need two more people to complete our team and it doesn't matter if it's a boy or a girl. Unfortuantelly, this method isn't somewhere incorrect.

The correct answer is 4280 so now I think there may be some extraneous selections in my method but I can't think of any. Would you please tell me where my approach falls?

The simple solution you give is ${20 \choose 4}-{8 \choose 4}-{12 \choose 4}$.
Another method would be ${8 \choose 3}{12 \choose 1}+ {8 \choose 2}{12 \choose 2} +{8 \choose 1}{12 \choose 3}$, giving the same correct result.
Your erroneous method is $3{8 \choose 3}{12 \choose 1}+ 4{8 \choose 2}{12 \choose 2} +3{8 \choose 1}{12 \choose 3}$, which illustrates the overcounting.
• Yes - those look correct to me. Hence the $3$s and $4$s multiplying in my last line – Henry Dec 14 '14 at 19:42